До тура: Ща покажу, кто тут главный!!!

После: Почему у меня меньше, чем у пятиклашек???

Главным образом олимпиада ориентирована на 8й класс, но мы рады видеть и более младших школьников :)

А ещё школьникам чуть проще чем вам, потому что у них есть возможность сдавать решения на частичные баллы.

-300?

Разумеется, нет :)

Раунд не случайно с ней пересекается.

for dislikers: soccer world cup final attracted 1.6 billion viewers while today's India vs Pak crikcet match expected to attract 1.5 billion viewers..so dislikers do respect other's choices as well.. although i will be going for the contest :P

div2 only

https://en.wikipedia.org/wiki/Mstislav_Keldysh

probably its him

So What? It is equal to say, "Although the time is not friendly for many other people, but the time is friendly for Chinese." Doesn't it?

No problem, rain will interupt the match

Как жаль что я 9-классник без интеллекта

No, I don't work together with him because he is quite nasty and rude.

wow your rating graph is inspirational!!

Just go to a reality where you are already master.

Statements are translated into English for the round.

Of course

Agree with you

Я подстраиваю свой far под себя

It'll be announced 3 seconds before the start of the round.

Obviously Contest

green Indians...

You can hack easy version only if you solved hard.

I found every 1 column flag and found out how wide they could become.
for every such flag, we get width*(width+1)/2 different possible flags.
Yes, |: 10^9 operations but it passed

Just use Java BigInteger with some cases handling

Big integer summation. use this. https://www.geeksforgeeks.org/sum-two-large-numbers/ I didn't even know how to implement summation .And, find out the index where s[i] != 0 near to the middle. then split the string and use that link for summation

Check this case

6 3
aaa
aab
bbb
bcd
xxx
xxx

The answer is 4.

And D = B.

I really think it feels right like that

because of the total number of submissions not huge, comparing for other contest

I saw some AC submitions with search in the BIT, maybe it helps you ^^

I used a Fenwick tree to get the n-th city, and it has passsed. https://mirror.codeforces.com/contest/1181/submission/55639812

There is no need to merge many Fenwick trees. Just maintain a simple sum-based seg tree.

I solved it in $$$O(q log(q) + m log(m) + q log(m))$$$ using Treap but finished the implementation when the contest was already finished :( 55644594

Let's solve the problem recursively. If $$$n=1$$$, the answer is "YES". Otherwise, we have to find a horizontal or vertical line which separates rectangles into smaller ones without intersecting any of them. If no such line exists, the answer is "NO". If it exists, we can greedily assume that it was the line of the last merge and solve problems recursively.

So, $$$O(n^2)$$$ (maybe $$$O(n^2 \cdot \log(n))$$$) is easy. How to speed it up? We'll try something like "smaller to greater", but in reverse. As we'll split the problem into two smaller ones, we can do it in complexity $$$O(size\_of\_the\_smaller\_part\_after\_the\_split)$$$ (possibly multiplied by $$$O(\log(n))$$$) and the overall complexity will be good. The problem is that we don't know if the line that we are looking for will be horizontal or vertical and we don't know if it'd be closer to the beginning or to the end of the list of sorted rectangles.

The solution is to try every option at the same time. Maintain $$$4$$$ sets, one for each possibility, go through each of them and if you'd find a good line, stop here, split each set into two smaller ones and the complexity will amortize.

I pre solved it in an offline manner.

1. Sort all the queries in terms of time stamp.
2. Interate through all the queries and merge the candidate cities in the process. I used two pointer like approach and fenwick tree to maintain this.

You could read my submission first if it passes. I am currently away from my laptop.

You should study some accepted submissions and learn simpler ways to approach the implementation. Then you can stop hating strings!

Here is my B. ~40 lines, 15 min.

String manipulation is much less annoying in python (in my opinion). My solution was less than 30 lines (and wasn't really optimized), so you might consider using a higher level language. Edit: 13 lines after cleaning up my code a little.

Such problems are easier to implement using languages which already have built-in support for large integers, like Python and Java.
It is always handy to learn the basics of such language (assuming you use C++).
55623625 Python solution in 8 lines.

May be because you were using += and concatenating strings. Use string.substr function instead. https://mirror.codeforces.com/contest/1181/submission/55632022

I'm not rated high enough to give an expert opinion about this contest, but from my perspective, this was a pretty lame contest. One or two more sentences in each problem to clear some stuff up would go a long way to helping us solve these problems, I feel like some of the information in the tasks were withhold from us just so the tasks would be harder to solve.